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Apr 26, 2021 at 10:36 vote accept Sascha
Feb 11, 2021 at 1:11 comment added Migalobe Sorry, I'm a bit confused. If $T(t)$ on $L^2(\mathbb{R}^n)$ is strongly continuous, this wouldn't contradict the non uniqueness of solutions?
May 30, 2018 at 11:46 answer added Bazin timeline score: 0
May 29, 2018 at 16:09 comment added Jochen Glueck Does Theorem 2 in my answer below address your intention of getting a general understanding of the situation, or are you more interested in semigroups whose generator does not have compact resolvent (as, e.g., the heat semigroup on the entire space $\mathbb{R}^n$ which is your motivating example).
May 27, 2018 at 18:51 answer added Jochen Glueck timeline score: 2
May 23, 2018 at 17:36 comment added Jochen Glueck Well, if we would like to be bold, we could conjecture that every subspace $X$ with the desired property is a direct sum of a subspace of $D(A)$ (which is, in addition, closed in $X$) and a finite-dimensional space. Unfortunately though, experience thaught me better not to be bold...
May 23, 2018 at 17:14 comment added Sascha @JochenGlueck thank you, that's right. However, I introduced these two examples only to illustrate what is possible and that it is not clear to me what will happen for arbitrary spaces.
May 23, 2018 at 16:08 comment added Jochen Glueck If $X\subset D(A)$, then the semigroup restricted to $X$ is uniformly continuous; this follows by applying the uniform boundedness principle to the restriction of the differential quotient $(T(t) - I)/t$ to $X$. This observation includes your spectral measure construction as a special case, but it does not include the case where $X$ is finite dimensional.
May 23, 2018 at 14:35 history edited Sascha CC BY-SA 4.0
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May 23, 2018 at 11:02 history asked Sascha CC BY-SA 4.0